Let A=[aᵢⱼ] be a matrix of order 3 × 3, with aᵢⱼ=(√2)ⁱ⁺ʲ. If the sum of all the elements in the third row of A² is α+β √2, α, β ∈ Z, then α+β is equal to :

Solution

Maths sir · 4 steps

Every entry splits as (√2)ⁱ (√2)ʲ, so A is a column times a row and A² = 14A.

1. Spot the structure. Since aᵢⱼ = (√2)ⁱ (√2)ʲ, take the column u = (√2; 2; 2√2). Then A = u uᵀ, a column times a row.

   A² = u (uᵀu) uᵀ = (uᵀu) A, uᵀu = 2 + 4 + 8 = 14

2. Third row of A. Row 3 of A is 2√2 × (√2, 2, 2√2) = (4, 4√2, 8). Its sum is 12 + 4√2.
3. Scale by 14. Because A² = 14A, row 3 of A² sums to 14(12 + 4√2) = 168 + 56√2. So α = 168 and β = 56.
4. So. α + β = 168 + 56 = 224, which is option (c).

The idea to keep. When aᵢⱼ = f(i) g(j), the matrix is a column times a row, so A² = (trace of A) A. Spot this and you skip the full multiplication.

Answer from the NTA final answer key. Solution by PrepWiser.